21 thoughts on “Permutations and Combinations – 5 Card Poker Hands”

Was here at the beginning of the year, now I’m back for my finals. Thanks
for this.

I’m applying this to yugioh, I need a consistent deck to win.

thank you very much this is a good explanation. had a problem in ap stats
asking to explain the probablility of 3 of a kind and you explained it
perfectly. thanks again

To give some more examples, (13C3) means to grab any three ranks from the
13 available ranks. So (13C3)(4C2)(4C1)(4C3) means grab any three ranks,
then choose 2 of the 4 cards for the first rank, then 1 of the 4 cards for
the second rank, and 3 of the 4 cards from the third rank. So how you set
it up depends on what you’re hand looks like. That’s why your original
examples were set up differently. Just to make sure, rank means a type of
card. Ace is a rank, 2 is a rank, 3 is a rank, etc.

13C1 is just used to represent you choosing one rank (type of card). But it
could be any rank. For example, (13C1) means it could be K, 10, Ace, 5,
Queen, 2, etc. So (13C1)(4C1) means to choose a rank, and then one card
from that rank. (13C1)(4C2) it means a pair from any rank. (13C1)(4C3)
means choose a rank and then choose 3 cards from that rank. But (13C1)
isn’t always used. For example, two pair uses (13C2)(4C2)(4C2). (13C2)
means pick any two ranks from the 13 available ranks.

Thank you so much for the clearing, I thought I loved math! I feel like a
child when it comes to permutation and combinations. So, 13C1 could only be
used if we are asked for a pair, 2 pairs, 3 or four of a kind, full house,
and royal flush?

The second example is a very specific type of 3 of a kind. It won’t have
the (13C1) because your 3 of a kind are JACKS. Plus your 4th card is fixed
as a Queen. This is why there is no (12C2) anymore from my example. So
(4C1) for the queen, (4C3) for the 3 Jacks, (44C1) to find the last card
from the remaining 44 ranks that is not a Queen or Jack. Hope this helps.
Just remember your examples are not general 3 of a kinds. Your formulas for
them will change because some cards are fixed.

These two examples don’t fit the general three of a kind scenario. Exactly
three kings means the other two cards could be the same (but not Kings).
This means you’re including some combinations that are a full house. Plus
since your 3 of a kind are KINGS, you don’t have the (13C1) at the
beginning. (13C1) allows for the 3 of a kind to be any card (3 Kings, 3
10s, 3 Aces, etc).

Hi Brian, my book solves (how many different 5-hand cards are possible that
consist of exactly 3 kings?) as (4C3)(48C2)=4512. Or, (how many different
5-hand cards are possible that consist of 1 Queen and 3 Jacks?) as
(4C1)(4C3)(44C1)=704 How are these different than your way of solving three
of a kind @ 9:07

Title says permutations and combinations but it’s just combinations

It’s the most common mistake, one I made the first time tried to do this
problem. If you calculate it your way, you’ll get twice as many
combinations. It’s because of the (13C1)*(12C1) part. If I remember
correctly, this means the order of the two ranks matter. (KK994 is
different than 99KK4) We know they are considered the same hand. It’s why
your number is twice as much as it should be. Instead, use (13C2). This
allows you to select 2 ranks, where order does not matter. KK994 = 99KK4

I tried the two pair example before he said the answer and I got
13C1*4C2*12C1*4C2*11C1*4C1. Why is that wrong? Or, what do I misunderstand
if I got that answer?

Good pace. Thank you.

Great way to explain everything!! Your a life saver!

really help me out with my HW！

This video helped me out a great deal with my HW. Thank you!

Thanks for the videos. They were a huge help!

GREAT VIDEOS, THANK YOU!

Your 3 of a kind example is incorrect. Should be 13C1x4C3x12C1x4C1x11C1x4C1

12C1 x 4C1 x 11C1 x 4C1 = 48 x 44 is more like a permutation because if you
switch the last two cards, this will treat it as a new hand. For example,
the way you have it written, QQQ5J would be a different hand than QQQJ5. So
you have to take 12C1 x 4C1 x 11C1 x 4C1 and divide by 2 to account for the
order of the last two cards. (12C1) x (11C1) / 2 = (48 x 44) / 2 is the
same as 12C2.

You broke down the steps very well and I understood everything. Thank you
so much…..

Was here at the beginning of the year, now I’m back for my finals. Thanks

for this.

I’m applying this to yugioh, I need a consistent deck to win.

thank you very much this is a good explanation. had a problem in ap stats

asking to explain the probablility of 3 of a kind and you explained it

perfectly. thanks again

To give some more examples, (13C3) means to grab any three ranks from the

13 available ranks. So (13C3)(4C2)(4C1)(4C3) means grab any three ranks,

then choose 2 of the 4 cards for the first rank, then 1 of the 4 cards for

the second rank, and 3 of the 4 cards from the third rank. So how you set

it up depends on what you’re hand looks like. That’s why your original

examples were set up differently. Just to make sure, rank means a type of

card. Ace is a rank, 2 is a rank, 3 is a rank, etc.

13C1 is just used to represent you choosing one rank (type of card). But it

could be any rank. For example, (13C1) means it could be K, 10, Ace, 5,

Queen, 2, etc. So (13C1)(4C1) means to choose a rank, and then one card

from that rank. (13C1)(4C2) it means a pair from any rank. (13C1)(4C3)

means choose a rank and then choose 3 cards from that rank. But (13C1)

isn’t always used. For example, two pair uses (13C2)(4C2)(4C2). (13C2)

means pick any two ranks from the 13 available ranks.

Thank you so much for the clearing, I thought I loved math! I feel like a

child when it comes to permutation and combinations. So, 13C1 could only be

used if we are asked for a pair, 2 pairs, 3 or four of a kind, full house,

and royal flush?

The second example is a very specific type of 3 of a kind. It won’t have

the (13C1) because your 3 of a kind are JACKS. Plus your 4th card is fixed

as a Queen. This is why there is no (12C2) anymore from my example. So

(4C1) for the queen, (4C3) for the 3 Jacks, (44C1) to find the last card

from the remaining 44 ranks that is not a Queen or Jack. Hope this helps.

Just remember your examples are not general 3 of a kinds. Your formulas for

them will change because some cards are fixed.

These two examples don’t fit the general three of a kind scenario. Exactly

three kings means the other two cards could be the same (but not Kings).

This means you’re including some combinations that are a full house. Plus

since your 3 of a kind are KINGS, you don’t have the (13C1) at the

beginning. (13C1) allows for the 3 of a kind to be any card (3 Kings, 3

10s, 3 Aces, etc).

Hi Brian, my book solves (how many different 5-hand cards are possible that

consist of exactly 3 kings?) as (4C3)(48C2)=4512. Or, (how many different

5-hand cards are possible that consist of 1 Queen and 3 Jacks?) as

(4C1)(4C3)(44C1)=704 How are these different than your way of solving three

of a kind @ 9:07

Title says permutations and combinations but it’s just combinations

It’s the most common mistake, one I made the first time tried to do this

problem. If you calculate it your way, you’ll get twice as many

combinations. It’s because of the (13C1)*(12C1) part. If I remember

correctly, this means the order of the two ranks matter. (KK994 is

different than 99KK4) We know they are considered the same hand. It’s why

your number is twice as much as it should be. Instead, use (13C2). This

allows you to select 2 ranks, where order does not matter. KK994 = 99KK4

I tried the two pair example before he said the answer and I got

13C1*4C2*12C1*4C2*11C1*4C1. Why is that wrong? Or, what do I misunderstand

if I got that answer?

Good pace. Thank you.

Great way to explain everything!! Your a life saver!

really help me out with my HW！

This video helped me out a great deal with my HW. Thank you!

Thanks for the videos. They were a huge help!

GREAT VIDEOS, THANK YOU!

Your 3 of a kind example is incorrect. Should be 13C1x4C3x12C1x4C1x11C1x4C1

12C1 x 4C1 x 11C1 x 4C1 = 48 x 44 is more like a permutation because if you

switch the last two cards, this will treat it as a new hand. For example,

the way you have it written, QQQ5J would be a different hand than QQQJ5. So

you have to take 12C1 x 4C1 x 11C1 x 4C1 and divide by 2 to account for the

order of the last two cards. (12C1) x (11C1) / 2 = (48 x 44) / 2 is the

same as 12C2.

You broke down the steps very well and I understood everything. Thank you

so much…..